Consider the integrand function:
x^3/(x^2-x-2)x3x2−x−2
Factorize the denominator:
x^2-x-2 = (x+1)(x-2)x2−x−2=(x+1)(x−2)
Now add ans subtract 11 to the numerator:
x^3/(x^2-x-2) = (x^3+1-1)/((x+1)(x-2)) = (x^3+1)/((x+1)(x-2)) -1/((x+1)(x-2))x3x2−x−2=x3+1−1(x+1)(x−2)=x3+1(x+1)(x−2)−1(x+1)(x−2)
Use the factorization: (x^3+1) = (x+1)(x^2-x+1)(x3+1)=(x+1)(x2−x+1) so that we can simplify:
(1) x^3/(x^2-x-2) = ( (x+1)(x^2-x+1) )/((x+1)(x-2)) -1/((x+1)(x-2)) = (x^2-x+1) /(x-2) -1/((x+1)(x-2)) (1)x3x2−x−2=(x+1)(x2−x+1)(x+1)(x−2)−1(x+1)(x−2)=x2−x+1x−2−1(x+1)(x−2)
Now note that:
x^2-x+1 = x^2-x-2 +3 = (x-2)(x+1) +3x2−x+1=x2−x−2+3=(x−2)(x+1)+3
so :
(x^2-x+1) /(x-2) = ((x-2)(x+1) +3)/(x-2) = x+1 +3/(x-2)x2−x+1x−2=(x−2)(x+1)+3x−2=x+1+3x−2
and that using partial fractions:
1/((x+1)(x-2)) = A/(x+1) +B/(x-2)1(x+1)(x−2)=Ax+1+Bx−2
1/((x+1)(x-2)) = (Ax-2A+Bx+B)/((x+1)(x-2))1(x+1)(x−2)=Ax−2A+Bx+B(x+1)(x−2)
{(A+B = 0),(-2A+B = 1):}
{(A = -1/3),(B = 1/3):}
Substitute this in (1) to have:
x^3/(x^2-x-2) =x+1 +3/(x-2) +1/(3(x+1))-1/(3(x-2))
x^3/(x^2-x-2) =x+1 +1/(3(x+1))+8/(3(x-2))
We can now integrate:
int x^3/(x^2-x-2)dx = x^2/2+x +1/3ln abs(x+1) +8/3 ln abs (x-2) +C
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